Question

2360. Longest Cycle in a Graph

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Description

You are given a directed graph of n nodes numbered from 0 to n - 1, where each node has at most one outgoing edge.

The graph is represented with a given 0-indexed array edges of size n, indicating that there is a directed edge from node i to node edges[i]. If there is no outgoing edge from node i, then edges[i] == -1.

Return _the length of the longest cycle in the graph_. If no cycle exists, return -1.

A cycle is a path that starts and ends at the same node.

 

Example 1:

Input: edges = [3,3,4,2,3]
Output: 3
Explanation: The longest cycle in the graph is the cycle: 2 -> 4 -> 3 -> 2.
The length of this cycle is 3, so 3 is returned.

Example 2:

Input: edges = [2,-1,3,1]
Output: -1
Explanation: There are no cycles in this graph.

 

Constraints:

• n == edges.length
• 2 <= n <= 105
• -1 <= edges[i] < n
• edges[i] != i

Solutions

Solution 1: Traverse Starting Points

We can traverse each node in the range $[0,..,n-1]$. If a node has not been visited, we start from this node and search for adjacent nodes until we encounter a cycle or a node that has already been visited. If we encounter a cycle, we update the answer.

The time complexity is $O(n)$ and the space complexity is $O(n)$, where $n$ is the number of nodes.

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class Solution:
  def longestCycle(self, edges: List[int]) -> int:
    n = len(edges)
    vis = [False] * n
    ans = -1
    for i in range(n):
      if vis[i]:
        continue
      j = i
      cycle = []
      while j != -1 and not vis[j]:
        vis[j] = True
        cycle.append(j)
        j = edges[j]
      if j == -1:
        continue
      m = len(cycle)
      k = next((k for k in range(m) if cycle[k] == j), inf)
      ans = max(ans, m - k)
    return ans

class Solution {
  public int longestCycle(int[] edges) {
    int n = edges.length;
    boolean[] vis = new boolean[n];
    int ans = -1;
    for (int i = 0; i < n; ++i) {
      if (vis[i]) {
        continue;
      }
      int j = i;
      List<Integer> cycle = new ArrayList<>();
      for (; j != -1 && !vis[j]; j = edges[j]) {
        vis[j] = true;
        cycle.add(j);
      }
      if (j == -1) {
        continue;
      }
      for (int k = 0; k < cycle.size(); ++k) {
        if (cycle.get(k) == j) {
          ans = Math.max(ans, cycle.size() - k);
          break;
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  int longestCycle(vector<int>& edges) {
    int n = edges.size();
    vector<bool> vis(n);
    int ans = -1;
    for (int i = 0; i < n; ++i) {
      if (vis[i]) {
        continue;
      }
      int j = i;
      vector<int> cycle;
      for (; j != -1 && !vis[j]; j = edges[j]) {
        vis[j] = true;
        cycle.push_back(j);
      }
      if (j == -1) {
        continue;
      }
      for (int k = 0; k < cycle.size(); ++k) {
        if (cycle[k] == j) {
          ans = max(ans, (int) cycle.size() - k);
          break;
        }
      }
    }
    return ans;
  }
};

func longestCycle(edges []int) int {
  vis := make([]bool, len(edges))
  ans := -1
  for i := range edges {
    if vis[i] {
      continue
    }
    j := i
    cycle := []int{}
    for ; j != -1 && !vis[j]; j = edges[j] {
      vis[j] = true
      cycle = append(cycle, j)
    }
    if j == -1 {
      continue
    }
    for k := range cycle {
      if cycle[k] == j {
        ans = max(ans, len(cycle)-k)
        break
      }
    }
  }
  return ans
}

function longestCycle(edges: number[]): number {
  const n = edges.length;
  const vis = new Array(n).fill(false);
  let ans = -1;
  for (let i = 0; i < n; ++i) {
    if (vis[i]) {
      continue;
    }
    let j = i;
    const cycle: number[] = [];
    for (; j != -1 && !vis[j]; j = edges[j]) {
      vis[j] = true;
      cycle.push(j);
    }
    if (j == -1) {
      continue;
    }
    for (let k = 0; k < cycle.length; ++k) {
      if (cycle[k] == j) {
        ans = Math.max(ans, cycle.length - k);
        break;
      }
    }
  }
  return ans;
}